Converting Fractions and Decimals

Date: 02/06/97 at 20:54:37
From: Doctor Keith
Subject: Re: Fractions and decimals
Hello!
In these explanations, if I make a step that is not obvious to you,
write back and I will clarify.
Question 1:
-----------
To figure this one out, you will want to remember what 3/5 means. It
means divide 3 by 5:
0 . 0 . 6
--- ------- -------
5/ 3 => 5/ 3 . 0 => 5/ 3 . 0
-0 -0
------- -------
3 . 0 3 . 0
-3 . 0
-------
0
Thus 3/5 = 0.6 which is the decimal expansion.
Another way to do this is to try to get the denominator into a power
of 10 (i.e.,: 1,10,100,1000,...). To do this, we first note that
2*5 = 10, so we need to multiply the denominator (bottom) by 2:
3 3 3 2 3*2 6
--- = --- * 1 = --- * --- = ----- = ----
5 5 5 2 5*2 10
Now we use the fact that dividing by ten shifts the decimal to the
left one place to find:
6/10 = 0.6
Both methods get the same answer. The first method will always be
about the same amount of work. The second method is fast if the
denominator is easy to change, but something like 3/41 would be more
trouble than it is worth.
Question 2
-----------
This is the same thing in reverse. We need to get the decimal into a
fraction and then cancel common factors. We note that 0.15 = 15/100,
from method 2 above. We need to get the factors of 15 and 100. Note
that the factors must be prime numbers (a prime number only has itself
and 1 as factors). Factors are whole numbers which will divide a
whole number into a whole number without any remainder. An example
best illustrates this:
6 = 2*3 = 6*1 Thus 1,2,3,6 are factors of 6
3 = 3*1 Thus 1,3 are factors of 3 and 3 is prime
Now we said we will want to get a prime factorization of a number,
which means we want to express the number as the product of its
smallest factors which are prime numbers, not 1:
6 = 2*3
12 = 2*2*3
Our factors are:
15 = 3*5
100 = 2*2*5*5
Now we just cancel the common factors (one 5 in this case):
15 3*5 3 3
----- = --------- = ------- = ----
100 2*2*5*5 2*2*5 20
3/20 is your answer. Note that I am going into a lot of detail,
trying to show you why you need to do things. In general you will
only need to do the last line, which actually is the work in the
problem.
Hope this helps out. Let us know if we can be of any further
assistance.
-Doctor Keith, The Math Forum
Check out our web site! http://mathforum.org/dr.math/